Chemistry Test-11

When the temperature of the endothermic reaction is increased, the value of the equilibrium constant increases.

However, when the temperature of the exothermic reaction is increased, the value of the equilibrium constant decreases.

So, With an increase in temperature, the value of equilibrium constant may increase or decrease.

The zeolite catalyst has sites that can remove hydrogen from an alkane together with the two electrons which bound it to the carbon. That leaves the carbon atom with a positive charge. Ions like this are called carbonium ions (or carbocations). The reorganization of these leads to the various products of the reaction.

Thus, the zeolite used as a catalyst in petrochemical industries for cracking of hydrocarbons and isomerization is ZSM-5.

Lone pair-lone pair repulsion is the most powerful repulsion depending upon the no. of lone pair present in the molecule.

Formula to calculate the number of lone pair:

Number of lone pairs on central atom =\(\frac{(\text{Total no. of valence shell electrons on central atom - No. of shared electrons by the central atom})} { 2}\)

The more the number of lone pairs in the molecule more will be the lone-pair lone pair repulsion.

\(\mathrm{XeF}_2\):

• Total no. of valence electrons \(=8(\mathrm{Xe})+7(\mathrm{~F})+7(\mathrm{~F})=22\) electrons
• Total no. of valence electron on \(\mathrm {Xe}=8\)
• Number of valence electron participating in bonding \(=2\)
• Leftover electron \(=6\)
• Total no. of lone pair on \(\mathrm {X e}=6 \div 2=3\)

For sample 1 we can write,

\(N _1= N _{01} e ^{-\lambda_1 t }\)

\(N _2= N _{02} e ^{-\lambda_2 t}\)

Given that at a particular time \(t N_1=2 N _2\)

Also given that for sample 1

\(A _1=-\frac{ dN }{ dt }= N _{01} \lambda_1 e ^{-\lambda_1 t }=5 \mu C _{ i }\)

\(A _2=-\frac{ d N _2}{ dt }= N _{02} \lambda_2 e ^{-\lambda_2 t }=10 \mu C _{ i }\)

Solving above equations we get,

\(\frac{\lambda_1}{\lambda_2}=\frac{1}{4}\)

\(T _{ S _1}=4 T _{ S _2}\)

\(\Rightarrow T _{ S _1}=20\) years and \(T _{ S _2}=5\) years is the possible answer.

\(\therefore\) Half-lives can be \(20\) years and \(5\) years, respectively.

Aspirin is not a habit-forming analgesic.

Heroin, Morphine and Codeine are all (potential) opiad drugs which interfere with the working of the central nervous system and hence can be habit-forming. Aspirin is an analgesic used to relieve minor aches and pains, as an antipyretic to reduce fever.

Various classes of drugs serve as analgesics, like paracetamol, non-steroidal anti-inflammatory drugs and opioid drugs. Opioid drugs are very effective as analgesics as they interfere with the working of the nervous system but can also cause sedation, nausea, euphoria, and respiratory depression.

Anhydrous CaCl₂ placed in the atmosphere saturated with water-it is a bulk phenomenon(absorption throughout the CaCl₂). It is not adsorption. Silica gel placed in the atmosphere saturated with water will adsorb water it is an example of physical adsorption.

The forces of van der Waals are weak intermolecular forces, depending on the distance between atoms or molecules. These forces emerge from the interactions between atoms/molecules that are not charged.

For example, Van der Waals forces arise from the fluctuation in the polarization of two near-to-neighboring particles.

III. In carbonate ion. \(C O_{3}^{2-}\) all three \(C-O\) bonds are identical due to resonance,

IV. \(P b^{2+}\) is more stable than \(P n^{4+}\) due to the inert pair effect, hence, prefers to form divalent compounds.

Given,

\(Cu ^{2+}+ e \rightarrow Cu ^{+} ;-\Delta G _1^{\circ}= E _1^{\circ} F \times 1\quad\dots(1)\)

\(Cu ^{+}+ e \rightarrow Cu ;-\Delta G _2^{\circ}= E _2^{\circ} F \times 1\quad\dots(2)\)

By adding the equation (1) and (2),

\(Cu ^{2+}+2 e \rightarrow Cu ;\)

\(-\Delta G _3^{\circ}=-\left[\Delta G _1^{\circ}+\Delta G _2^\circ\right]=-[0.15 F +0.5 F ]\)

or \(2 \times E _3^{\circ} F =0.65 F\)

(as \(\Delta G =- nF E ^{\circ}\) )

\(\therefore E _3^0=0.325 V\)

So for the reverse reaction.

\(\therefore Cu \rightarrow Cu ^{2+}\)

\(2 eE ^{\circ}=-0.325 V\)

Given -

Energy per fission of \({ }_{92} U ^{235}\) nuclei \(=200 MeV =200 \times 1.6 \times 10^{-13}=3.2 \times 10^{-11}\) joule,

The total energy required \((E)=1\) Megawatt \(=10^{6}\) joule / second

The number of atoms disintegrated in a day is given by \(\Rightarrow\) Number of atoms \(=\frac{P t}{E}=\frac{24 \times 36 \times 10^{8}}{32 \times 10^{-12}}=\frac{864 \times 10^{8}}{32 \times 10^{-12}}=27 \times 10^{20}\)

The number of atoms disintegrated, \(N=27 \times 10^{20}\)

The amount of mass-consumed is given by

\(\Rightarrow\) Mass consumed \(=\frac{N}{N_{A}} \times M\)

Substituting the values \(N_{A}=6.023 \times 10^{23}\) and \(M=235\)

\(\Rightarrow\) Mass consumed \(=\frac{27 \times 10^{20}}{6.023 \times 10^{23}} \times 235=1.05\) gm